LeetCode-in-Java

287. Find the Duplicate Number

Medium

Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.

There is only one repeated number in nums, return this repeated number.

You must solve the problem without modifying the array nums and uses only constant extra space.

Example 1:

Input: nums = [1,3,4,2,2]

Output: 2

Example 2:

Input: nums = [3,1,3,4,2]

Output: 3

Example 3:

Input: nums = [1,1]

Output: 1

Example 4:

Input: nums = [1,1,2]

Output: 1

Constraints:

• 1 <= n <= 105
• nums.length == n + 1
• 1 <= nums[i] <= n
• All the integers in nums appear only once except for precisely one integer which appears two or more times.

Follow up:

• How can we prove that at least one duplicate number must exist in nums?
• Can you solve the problem in linear runtime complexity?

Solution

public class Solution {
    public int findDuplicate(int[] nums) {
        int[] arr = new int[nums.length + 1];
        for (int num : nums) {
            arr[num] += 1;
            if (arr[num] == 2) {
                return num;
            }
        }
        return 0;
    }
}

Time Complexity (Big O Time):

The time complexity of this program is O(n), where ‘n’ is the length of the input array nums. This is because the program iterates through the entire array exactly once in a single pass. In the worst case, it might need to iterate through the entire array to find the duplicate.

Space Complexity (Big O Space):

The space complexity of the program is O(n), where ‘n’ is the length of the input array nums. The program creates an integer array arr of size nums.length + 1, which requires additional space. The size of this array is directly proportional to the size of the input array nums.

In summary, the provided program has a time complexity of O(n), where ‘n’ is the length of the input array nums, and it has a space complexity of O(n), indicating linear space usage.

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